Can't exit a do/while loop - please help

Hello again,

I was wondering if anyone might be able to tell me why the program won't exit the loop when response equals 'q' or 'Q'. When I select a Q or a q, the program just loops contiuously. Any help would be appreciated.

Thanks
Vireyda



[CODE]
do{
printf("Please select a transaction:\n");
printf("Cheque - C or c\nDeposit - D or d\nPrint - P or p\nQuit - Q or q\n");
fflush(stdin);
scanf("%c", &response);
switch(response)
{case'c':case'C':
Cheque(balance);
break;
case 'd': case'D':
Deposit(balance);
break;
case 'p': case'P':
Print(balance);
break;
case 'q': case'Q':
break;
default:
printf("That is an invalid selection. Please try again.\n ");
}
}while(response!='q'||response!='Q');
[\CODE]

You want to quit the loop when
(response == 'q') || (response == 'Q')

That means you want to continue as long as

(response != 'q') && (response != 'Q')

Fundamental boolean logic.

Try it.

Dave

I once had this problem a while back, and i also posted it here. The answer i got was instead of using the OR ( || ) operator, i should have used the AND ( && ) operator. I sat down thinking how ilogical this sounds, and coudn't come up with an answer. It still annoys me today whenever i think about it ( So if anyone would like to answer this question, i would be gratfull).

Anyways, use the && operator instead of the || operator in your test expression and it should work fine.

Woohoo! Listen to Evans! thanks Evans! answered my quetison too

Code:

---

while(response!='q'||response!='Q');

---

if you look at the logic of the statment

(response!='q') is true(1) when any other letter is entered and false(0) when you enter 'q'

(response!='Q') is true(1) when any other letter is entered and false(0) when you enter 'Q'

one is always going to be true and the other false when you try to quit

when you OR 1 and 0 you get 1 which is what you are doing so it evaluates to while(1) which is an infinite l oop

when you AND 1 and 0 you get 0 which will break the loop and exit so just change the || to && and you should be alright.

?? :confused: Wait....i still don't get it. Why won't
response=='q' || response 'Q'
exit the loop?

Quote:

---

Originally posted by Dave Evans
You want to quit the loop when
(response == 'q') || (response == 'Q')

That means you want to continue as long as

(response != 'q') && (response != 'Q')

Fundamental boolean logic.

Try it.

Dave

---

I dont think it's a matter of boolean logic because it is not logic at all that he will write q and Q simultaneously

but i think he wrote on his code

Code:

---

while(response!='q'||response!='Q');

---

i suggest that he will replace it by

Code:

---

while((response!='q')||(response!='Q'));

---

or

Code:

---

while(!((response=='q')||(response=='Q')));

---

i am not sure anyway .. u might be right

Quote:

---

Originally posted by jrahhali
?? :confused: Wait....i still don't get it. Why won't
response=='q' || response 'Q'
exit the loop?

---

Because the loop continues as long as the stuff in the parentheses is true. (Therefore the exit condition is when the stuff is not true.)

Code:

---

do {

} while(ThisStuffIsTrue);

---

Quote:

---

Originally posted by M_Ghani
I dont think it's a matter of boolean logic because it is not logic at all that he will write q and Q simultaneously

but i think he wrote on his code

Code:

---

while(response!='q'||response!='Q');

---

i suggest that he will replace it by

Code:

---

while((response!='q')||(response!='Q'));

---

or

Code:

---

while(!((response=='q')||(response=='Q')));

---

i am not sure anyway .. u might be right

---

Well, let's simplify the notation:

I didn't mean to confuse anyone with jargon, like "Boolean logic", but one the things you learn is (in C-language notation):

The following is always true:


!(x || y) == (!x && !y)

It's not a "real" proof, but maybe you can convince yourself it's worth considering if you try to evaluate each with all possible values of x and y. You can do this in your head, on pencil and paper, or in a C program:

x = false, y = false.
x = false, y = true.
x = true, y = false.
x = true, y = true.

This topic comes up all of the time. Once you "get it", it's yours for life.



Regards,

Dave

Quote:

---

Originally posted by Dave Evans
Well, let's simplify the notation:

I didn't mean to confuse anyone with jargon, like "Boolean logic", but one the things you learn is (in C-language notation):

The following is always true:


!(x || y) == (!x && !y)

It's not a "real" proof, but maybe you can convince yourself it's worth considering if you try to evaluate each with all possible values of x and y. You can do this in your head, on pencil and paper, or in a C program:

x = false, y = false.
x = false, y = true.
x = true, y = false.
x = true, y = true.

This topic comes up all of the time. Once you "get it", it's yours for life.



Regards,

Dave

---

yeah .. I got it

so that means that

Code:

---

while(!((response=='q')||(response=='Q')));

---

is correct and is the same as

Code:

---

while((response!='q')&&(response!='Q'));

---

Thanks everyone. Your help was invaluable. The program works for me now and is a big relief off my shoulders.

Vireyda :)

Thankyou so much for the logic!! That IS logical when you use bool- it also helped me with a program!!